lc-257

给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [1,2,3,null,5]
输出：["1->2->5","1->3"]

示例 2：

输入：root = [1]
输出：["1"]

提示：

• 树中节点的数目在范围 [1, 100] 内
• -100 <= Node.val <= 100

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        List<TreeNode> curList = new ArrayList<>();
        if (root == null) {
            return result;
        }
        curList.add(root);
        getPath(result, curList);
        return result;
    }

    private void getPath(List<String> result, List<TreeNode> curList) {
        TreeNode node = curList.get(curList.size() - 1);
        if (node.left == null && node.right == null) {
            String str = "";
            for (TreeNode n : curList) {
                if (str.length() == 0) {
                    str = str + n.val;
                } else {
                    str = str + "->" + n.val;
                }
            }

            result.add(str);
            return;
        }

        if (node.left != null) {
            curList.add(node.left);
            getPath(result, curList);
            curList.remove(curList.size() - 1);
        }
        if (node.right != null) {
            curList.add(node.right);
            getPath(result, curList);
            curList.remove(curList.size() - 1);
        }
    }
}
//leetcode submit region end(Prohibit modification and deletion)